jeffrieskids13
12.12.2019 •
Chemistry
How many grams of sodium chloride are present in a 0.75 m solution with the volume of 500.0 milliliters
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Ответ:
V: 500mL = 0,5L
C = n/V
n = C×V
n = 0,75×0,5
n = 0,375 mol
mNaCl: 23+35,5 = 58,5 g/mol
58,5g 1 mol
Xg 0,375 mol
X = 58,5×0,375
X = 21,9375g NaCl
:•)
Ответ:
So the Atomic Number 81 refers to the Element:
Thallium